# Binary Tree Cameras

Given a binary tree, we install cameras on the nodes of the tree.

Each camera at a node can monitor**its parent, itself, and its immediate children**.

Calculate the minimum number of cameras needed to monitor all nodes of the tree.

**Example 1:**

![](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png)

```
Input: 
[0,0,null,0,0]
Output: 
1
Explanation: 
One camera is enough to monitor all nodes if placed as shown.
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png)

```
Input: 
[0,0,null,0,null,0,null,null,0]
Output: 
2

Explanation:
 At least two cameras are needed to monitor all nodes of the tree. The above image shows one of the valid configurations of camera placement.
```

Note:

```
The number of nodes in the given tree will be in the range [1, 1000].
Every node has value 0.
```

分析

3个状态NOT\_MONITORED = 0

```
    MONITORED\_NOCAM = 1

    MONITORED\_WITHCAM = 2
```

尽量往上放

1 2个儿子都被监视，求助parent

2任意儿子不被监视，本地放置 res++

3 直接返回被监视（可能儿子处放置

```
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def minCameraCover(self, root: TreeNode) -> int:
        NOT_MONITORED = 0
        MONITORED_NOCAM = 1
        MONITORED_WITHCAM = 2
        cameras = 0
        if not root:
                return 0

        def dfs(root: TreeNode) -> int:
            nonlocal cameras
            if not root:
                return MONITORED_NOCAM
            left = dfs(root.left)
            right = dfs(root.right)

            if left == MONITORED_NOCAM and right ==MONITORED_NOCAM: 
                return NOT_MONITORED
            elif left == NOT_MONITORED or right == NOT_MONITORED:
                cameras += 1
                return MONITORED_WITHCAM            
            else:
                return MONITORED_NOCAM
        top = dfs(root)

        return cameras+1 if  top== NOT_MONITORED else cameras
```


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